Theorem difdif 4097

Description: Double class difference. Exercise 11 of [TakeutiZaring] p. 22. (Contributed by NM, 17-May-1998.)

Assertion

Ref	Expression
difdif	⊢ (𝐴 ∖ (𝐵 ∖ 𝐴)) = 𝐴

Proof of Theorem difdif

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		pm4.45im 840	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴)))
2		iman 406	. . . . 5 ⊢ ((𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴) ↔ ¬ (𝑥 ∈ 𝐵 ∧ ¬ 𝑥 ∈ 𝐴))
3		eldif 3923	. . . . 5 ⊢ (𝑥 ∈ (𝐵 ∖ 𝐴) ↔ (𝑥 ∈ 𝐵 ∧ ¬ 𝑥 ∈ 𝐴))
4	2, 3	xchbinxr 338	. . . 4 ⊢ ((𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴) ↔ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴))
5	4	anbi2i 634	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴)) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴)))
6	1, 5	bitr2i 279	. 2 ⊢ ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴)) ↔ 𝑥 ∈ 𝐴)
7	6	difeqri 4091	1 ⊢ (𝐴 ∖ (𝐵 ∖ 𝐴)) = 𝐴

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 400   = wceq 1567   ∈ wcel 2149   ∖ cdif 3910

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1570  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-v 3465  df-dif 3916

This theorem is referenced by:  dif0  4340  undifabs  4441